After having learnt the statement of Bent’s rule in the earlier post , let us now study the rule with some examples.

We know that *bond angle(θ) **∝ % s character* ( from post no 71) .Then , according to Bent’s rule ,

Now let us apply this to various examples and study Bent’s rule .

**Example 1 – Methyl Fluoride (CH _{3}F molecule) .**

In a methyl fluoride molecule , **fluorine** is the most **electronegative element**. Just like methane (CH_{4}) , this molecule is sp^{3} hybridised too.However, the distribution of s- character is not uniform in all hybrid orbitals and so bond angles change too.This change is due to the presence of F atom. We can predict the change in bond angles using Bent’s rule.

According to Bent’s rule, the orbital with less %s character will be directed towards the more electronegative element. So, we can conclude that , the bond between C- F has** less % s- character** compared to C-H bonds. Also, the bond angle will decrease. Experimentally the H-C-F bond angle is found to be 108.2º , which is less than 109.5º .

When the %s character in C- F bond decreases, it is accounted for by increase in the %s-character in the other three C-H bonds. So, C-H bonds have more %s character than expected ,resulting in increase in bond angle between them. Experimentally , the H-C-H bond is found to be 110.2º.

**Example 2 – Sulphur tetrafluoride (SF _{4}) .**

We were skeptical about the structure of this molecule in the earlier post.Let us study it’s structure using Bent’s rule. SF_{4} is sp^{3}d hybridized and has TBP geometry .There are four fluorine atoms and a lone pair in this molecule. As we know, according to Pauling scale of electronegativity , fluorine is the most electronegative element in the periodic table (refer post no 46) . In the TBP geometry , there are three equatorial positions and two axial positions.

We can **split sp ^{3}d as sp^{2} + pd. **

As seen above , the sp^{2} orbitals occupy the equatorial positions and the pd orbitals occupy the axial position. How can we be sure of this ? Let’s check whether this is true –

Experimentally , we find out that the angle between equatorial substituents is 120º. We also know,

i = – 1/ cosθ

∴i = – 1/ cos 120

i= – 1/ -(1/2) ..As cos 120 = -1/2

∴ i = 2

So, as the hybridization index is equal to 2 , the equatorial orbitals are sp^{2} orbitals.

Or we can use the formula from post 69 – **f _{s}=cosθ/(cosθ-1) **

∴fraction of s- character in equatorial positions with θ = 120º is ,

**f _{s}= cos(120) / (cos 120 -1)
= (-1/2) /(-1/2 -1)
=(-1/2)/ (-3/2)
= 1/3
=0.33.**

** **We know that it is sp^{2} orbitals which have 33% s- character , so the equatorial hybrid orbitals are sp^{2} hybridized.

lll^{ly }we know experimentally , that the angle with axial substituents is 90º. So,

**f _{s}=cosθ/(cosθ-1)
** = cos 90 / cos90 – 1

= 0/(0-1)

=0.

So, the amount of s- character in axial orbitals is zero i.e the pd part(no s character)occupies the axial position. Applying Bent’s rule –

So the correct structure for this compound is –

We shall study some more examples of this rule in our next post. Till then ,

Be a perpetual student of life and keep learning …

Good day !

Thank you so much sir for explaining the Bent rule in such a simple way. I was having little difficulty in it, but after reading your post I developed interest in this topic and in inorganic chemistry because my interest was always been in organic chemistry.

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I am so glad to hear from you …Thanks 😊

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